formulario Calculo

June 20, 2019 | Author: theband | Category: N/A
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formulario basico y util para calculo diferencial e integral...

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Formul ormular ario io de Prec Prec´ ´ alcu alculo lo.. 1.

5. Leyes de los logaritmos. a ) loga (P Q) = loga (P ) P ) + loga (Q) P  b ) loga = loga (P ) P ) loga (Q) Q

Los Numeros. u ´meros.



1. Leyes de los exponentes y radicales. a ) am an = am+n d )

a b



n

n

=

g ) a1/n =  j )  j )

b ) (am )n = amn

a bn

e)

√a

√ab = √a √b n

n

n

a = am−n n a

h ) am/n =

n

k )

  n

a = b

c ) loga (Qn ) = n loga (Q)

c ) (ab) ab)n = an bn

m

 f )  f ) a−n =

√am √a √b

e ) loga (ax ) = x



n

l )

n

 √

m

n

d ) aloga (x) = x

1 an

i ) am/n = ( n a)

n

a=

 f )  f ) loga (1) = 0

m

g ) aloga (a) = 1

√a

mn

h ) log(x log(x) = log10 (x)

2. Productos Productos Notables. Notables.

i ) ln(x ln(x) = loge (x)

− y) = x2 − y2 b ) Binomio al Cuadrado: (x (x ± y )2 = x2 ± 2xy + y 2 c ) Binomio al Cubo: (x (x ± y )3 = x3 ± 3x2 y + 3xy 3xy 2 ± y 3

a ) Binomios Binomios Conjugados: Conjugados: (x (x + y )(x )(x

 j )  j ) Cambi Cambioo de base base::

logb (Q) logb (a)

log loga (Q) =

2. Soluc Solucio ione ness Exact Exactas as de ecu ecuac acioiones Algebraicas

2

d ) d ) (x + y ) = x2 + 2 xy + y 2 e ) (x



− y)2 = x2 − 2 xy + y2 3

 f )  f ) (x + y ) = x3 + 3 x2 y + 3 xy 2 + y 3 g ) (x

6. Soluciones Exactas de Ecuaciones Algebraicas.

− y)3 = x3 − 3 x2y + 3 xy2 − y3

a ) La Ecuaci´ on on Cuadr Cu adr´ ´ atica ati ca:: ax2 + bx + c = 0 tiene soluciones: b b2 4ac x= 2a 2 El n´ umero umero b 4ac se llama discriminante de la ecuaci´ on. on. i) Si b2 4ac > 0 las ra´ ra´ıces son reales y diferentes. 2 ii) Si b 4ac = 0 las ra´ ra´ıces son reales e iguales. 2 iii) Si b 4ac < 0 las la s ra´ıces ıces son so n complejas compl ejas conjugaconjug adas. b ) Para la Ecuaci´ on o n C´ ubica: ubica: x3 + ax2 + bx + c = 0 sean:

4

h ) (x + y ) = x4 + 4 x3 y + 6 x2 y 2 + 4 xy3 + y 4 i ) i ) (x

− ±√ −

− y)4 = x4 − 4 x3y + 6 x2y2 − 4 xy3 + y4 5

 j )  j ) (x + y ) = x5 + 5 x4y + 10 x3y 2 + 10 x2y 3 + 5 xy4 + y 5



− y)5 = x5 − 5 x4y + 10 x3y2 − 10 x2y3 + 5 xy4 − y5 3. Teorema del Binomio. Sea n ∈ N, entonces: k ) k ) (x

n

n

(x + y ) =

  r=0

Nota:



− − −

n n−r r x y r

n n! = n Cr = r r!(n !(n r)!



Q=

3b

9

4. Factores Notables. a ) Diferencia Diferencia de Cuadrados: Cuadrados: x2

− y2 = (x ( x + y )(x )(x − y ) b ) Suma de Cubos: x3 + y 3 = (x + y)(x )(x2 − xy + y 2 ) c ) Diferencia Diferencia de Cubos: Cubos: x3 − y 3 = (x ( x − y )(x )(x2 + xy + y 2 ) d ) d ) Trinomio Cuadrado Perfecto: x2 ± 2xy+ xy + y 2 = (x± y)2 e ) x2 − y 2 = (x ( x − y ) (x + y ) 3− 3  f )  f ) x y = (x ( x − y ) x2 + xy + y 2 g ) x3 + y 3 = (x ( x + y ) x2 − xy + y 2 h ) x4 − y 4 = (x ( x − y ) (x + y ) x2 + y 2 i ) i ) x5 − y 5 = (x ( x − y ) x4 + x3 y + x2 y 2 + xy3 + y 4  j )  j ) x5 + y 5 = (x ( x + y ) x4 − x3 y + x2 y 2 − xy3 + y 4 k ) k ) x6 −y 6 = (x ( x − y) (x + y ) x2 + xy + y 2 x2 − xy + y 2 l ) l ) x4 + x2 y 2 + y4 = x2 + xy + y 2 x2 − xy + y 2 m ) x4 + 4 y 4 = x2 − 2 xy + 2 y 2 x2 + 2 xy + 2 y 2

             

     1

S  =

− a2 ,

R=

    3

R+

9ab

Q3 + R2 ,

54

T  =

Entonces las soluciones son: a x1 =S + S  + T  3 S + S  + T  a x2 = + + 2 3



x3 =

 −  −

S + S  + T  a + 2 3

− 27 27cc − 2a3

  −

  −   3

R

√ (S  − T ) T ) 3 2 √ (S  − T ) T ) 3 2

Q3 + R2

 

i i

El n´ umero umero Q3 +R2 se llama discriminante de la ecuaci´ on. on. i) Si Q3 + R2 > 0, hay una ra´ ra´ız real y dos son s on complejas conjugadas. ii) Si Q3 + R2 = 0, las ra´ ra´ıces son reales y por lo menos dos son iguales. iii) Si Q3 + R2 < 0, las l as ra r a´ıces son reales y diferentes. di ferentes.

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Formul ormular ario io de C´ alcu alculo lo.. Derivadas. En este formulario: k, c R son constantes reales, f  = f ( f (x), u = u(x) y v = v(x) son funciones que dependen de x.



Funciones Funcion es Trigono rig onom´ m´ etricas etr icas:: Funci´ on:

Su Derivada:

f  = sen(u sen(u)

f ′ = cos(u cos(u) u′

f  = cos(u cos(u)

f ′ =

f  = tan(u tan(u) f  = csc(u csc(u)

F´ ormulas ormul as B´ asicas asi cas:: Funci´ on:

Su Derivada:

f  = k

f ′ = 0

f  = sec(u sec(u) f  = cot(u cot(u)

·

− sen(u sen(u) · u′ f ′ = sec2 (u) · u′ f ′ = − csc(u csc(u) cot( cot(u u) · u′ f ′ = sec(u sec(u) tan( tan(u u) · u′ f ′ = − csc2 (u) · u′

Linealidad de la derivada: f  = k u

· f  = u ± v f  = k · u ± c · v

f ′ = k u′

Funciones uncion es Trigonom´ rigon om´ etricas etricas Inversas:

·

f ′ = u′

± v′ f ′ = k · u′ ± c · v′

Regla del Producto: f ′ = u v′ + v u′

f  = u v

·

·

·

Funci´ on: f  = arc sen( sen(u)

Su Derivada: u′ ′ f  = ; 1 u2

f  = arc cos( cos(u)

f ′ =

− √1u− u2 ; |u| < 1

f  = arctan(u arctan(u)

f ′ =

u′ 1 + u2

f  = arccsc(u arccsc(u)

f ′ = −

f  = arcsec(u arcsec(u)

f ′ =

u √ ; |u| > 1 u u2 − 1

f  = arccot(u arccot(u)

f ′ =

− 1 +u u2 ; |u| > 1

√ −

Regla del Cociente: Cociente: f  =

u v

f ′ =

v · u′ − u · v′ v2

Regla de la Cadena (Composici´ on on de funciones) f  = u(x) v(x)



f ′ = [u(v(x))]′ v′ (x)

·

Regla de la Potencia: f  = vn f  = k vn

·





u √ u u2 − 1





f ′ = n vn−1 v′

· · f ′ = k · n · vn−1 · v′

Funciones Exponenciales: u

|u| < 1

f  = e

f ′ = eu · u′

f  = au

f ′ = au ln(a ln(a) u′

·

Funciones Funcio nes Logar Log ar´ ´ıtmica ıtm icas: s:



·

Funciones Hiperb´ olicas: olicas: Funci´ on:

Su Derivada:

f  = senh(u senh(u)

f ′ = cosh(u cosh(u) u′

f  = cosh(u cosh(u) f  = tanh(u tanh(u) f  = csch(u csch(u)

· f ′ = senh(u senh(u) · u′ f ′ = sech2 (u) · u′ f ′ = −csch(u csch(u) coth( coth(u u) · u′

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Definici´ on on 1. Ecuac Ecuaci´ i´ on en Variables Separadas. on Consideremos la ecuaci´on on con forma est´andar: andar: M ( M (x)dx )dx + N ( N (y )dy )dy = 0

(1)

La soluci´ on se obtiene integrando directamente: on

 

M ( M (x)dx )dx +

 

N ( N (y )dy )dy = C 

Definici´ on on 2. Ecuac Ecuaci´ i´ on en Variables Separables. on Las siguientes dos ecuaciones, son ecuaciones en variables separables.

)dx + M 2 (x)N 2 (y )dy )dy = 0 M 1 (x)N 1 (y)dx

dy = f ( f (x)g (y ) dx

(2)

(3)

Para determinar la soluci´on on de la Ec.(2), se divide la ecuaci´on on entre: M 2 (x)N 1 (y), para reducirla a la ecuaci´ on en variables separadas: on

La soluci´ on de la Ec.(3), se obtiene al dividir entre on g (y) y multiplicar por dx d x, para reducirla a la ecuaci´on on en variables separadas:

M 1 (x) N 2 (y ) dx + dy = 0 M 2 (x) N 1 (y )

1 dy = f ( f (x)dx )dx g (y )

ahora s´olo olo se integra directamente:

  Definici´

M 1 (x) dx + M 2 (x)

3. Ec

 

N 2 (y ) dy = C  N 1 (y )

ci´ cion ´ on Lineal.

ahora s´olo olo se integra directamente:

 

1 dy = g (y )

 

f ( f (x)dx )dx + C 

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Definici´ on on 5. Ecuaciones Ecuaciones Exactas Exactas o en Diferenci Diferenciales ales Totales. Totales. Consideramos la ecuaci´on: on: M (x, y )dx )dx + N ( N (x, y )dy )dy = 0

(8)

donde se cumple: M y = N x . La soluci´on on se obtiene de calcular: i) ii)

u=

 

M (x, y )dx )dx,

calculamos: uy

 

iii)

v = [N ( N (x, y )

− uy ]dy ]dy

iv) iv)

La solu soluci ci´ on o ´n general impl´ıcita ıcita es: u + v = C 

Definici´ on on 6. Factor Integran Integrante. te. Consideremos la ecuaci´on: on: M (x, y )dx )dx + N ( N (x, y )dy )dy = 0

(9)

donde M y = N x . Para determinar la soluci´on on de esta ecuaci´on, on, se tiene que reducir a una ecuaci´on on exacta; exa cta; as´ı que q ue primero se debe calcular uno de los dos posibles factores integrantes:



M y N x dx N 

  1) µ(x) = e



N x My dy M 

  2) µ(y ) = e



segundo se multiplica la Ec.(9) por el factor integrante que exista y se obtiene la ecuaci´on exacta: exacta:

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B) Con el cambio de variable v = xy . La Ec.(11) Ec.(11) se reduce a la ecuaci´on on en variables separadas: dy M ( M (v, 1) + dv = 0 y N ( N (v, 1) + vM ( vM (v, 1)

la cual cual se inte integra gra dire direct ctam amen ente te

la soluci´ on de la Ec.(11) se obtiene al sustituir nuevamente v por on

x y

dy + y

M ( M (v, 1) dv = C  N ( N (v, 1) + vM ( vM (v, 1)

   

en el resultado de la integral.

La Ec.(12) Ec.(12) se reduce a la ecuaci´on on en variables separadas: dv 1 f (v,1) v, 1)

−v

=

dy y

la cual se integra directamente

 

1 f ( f (v,1) v,1)

la soluci´ on de la Ec.(12) se obtiene al sustituir nuevamente v por on

x y

y2

···

yn

−v

=

dy + C  y

 

en el resultado de la integral.

I. Wronskiano. y1

dv

Rengl´ on on de las funciones.

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Caso iii. Ra´ Ra´ıces Conjugadas Conjug adas Complejas, Complej as, y3 (x). Sean m1 = α1 β 1 i, m2 = α2 β 2 i, m3 = α3 β 3 i , . . . las ra´ ra´ıces complejas conjugadas de (15), entonces, otra parte de yh (x) se escribe como:

±

±

±

y3 (x) = eα1 x C 1 cos(β  cos(β 1 x) + C 2 sen(β  sen(β 1 x) +





eα2 x C 3 cos(β  cos(β 2 x) + C 4 sen(β  sen(β 2 x) +





eα3 x C 5 cos(β  cos(β 3 x) + C 6 sen(β  sen(β 3 x) +

Nota: Nota: Obs´ ervese ervese que se toma el valor positivo positivo de β  en todos las casos.





···

(18)

Caso iv. Ra´ Ra´ıces Conjuga C onjugadas das Complejas Co mplejas Repetidas, Repeti das, y4 (x). Sean m1 = α βi = m2 = α βi = m3 = α βi = las ra´ ra´ıces conjugadas complejas repetidas rep etidas de (15), entonces, otra parte de yh (x) se escribe como:

±

±

±

···

y4 (x) = eαx C 1 cos(βx cos(βx)) + C 2 sen(βx sen(βx)) +





eαx C 3 cos(βx cos(βx)) + C 4 sen(βx sen(βx)) +

x





2 x

eαx C 5 cos(βx cos(βx)) + C 6 sen(βx sen(βx)) +





···

(19)

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Segundo. Caso i. Ecuaci´ on on de segundo orden. La soluci´ solucion o´n particular tiene la forma: y p(x) = u1 y1 + u2 y2 donde: u′1 = u′2 =

−g(x)y2 ,

u1 =

g (x)y1 , W [ W [y1 , y2 ]

u2 =

W [ W [y1 , y2 ]

g (x)y2 dx W [ W [y1 , y2]

  −  

g (x)y1 dx W [ W [y1 , y2]

Caso ii. Ecuaci´ on on de tercer orden. La soluci´ solucion o´n particular tiene la forma: y p (x) = u1 y1 + u2 y2 + u3 y3 donde: ′



g (x)[y )[y2 y3 − y3 y2 ] u′1 = , W [ W [y y y ]

u1 =

 





g (x)[y )[y2 y3 y3 y2 ] dx W [ W [y y y ]



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• Primer Teorema de Traslaci´on on o de Desplazamiento: at f (t)} = F ( F (s − a) L  {e f ( Primero identificamos el valor de a y se calcula L  {f ( f (t)} = F ( F (s). Segundo se calcula F ( F (s) s=s−a , y as´ as´ı se cumple que at L  {e f ( f (t)} = F ( F (s − a). • Funci´on on Escal´on on Unitario Unitario de Heaviside, Heaviside, denotada denotada como U  (t − a) o H (t − a). 0, 0 ≤ t ≤ a; H (t − a) = U  (t − a) = 1, t ≥ a. • Funci´on on por partes en t´ erminos erminos la l a funci´on on escal´on on unitario. Sea f 1 (t) 0 ≤ t ≤ a f 2 (t) a ≤ t < b f ( f (t) = f 3 (t) b ≤ t < c f 4 (t) t ≥ c





entonces:



f ( f (t) = f 1 (t)U  (t) + f 2 (t)



 

− f 1(t) U  (t − a) +



f 3 (t)



− f 2(t) U  (t − b) +



f 4 (t)



− f 3(t) U  (t − c)

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